Waves in Pipe

7/29/2019 Waves in Pipe

1/8

Lecture 27 Page 1

PHYSICS 151 Notes for Onl ine Lecture #27Reflect ion of Waves

Waves obey something we call the principle of linear superposition. That is, if two waves arein the same region of

space at the same

time, they willinteract with each

other. Linear

superposition allowsus to describe how

they interact fairly

easily. If we were toplot the two waves as

a function of time,

they might look likethe top two waves in

the picture to the

right. Linearsuperposition means

that we just add the value of each wave and plot that as the sum. So the result of this

combination of waves is another sine wave, but with an amplitude that is the sum of theamplitudes of the two starting waves.

Lets repeat this, but shift one of the waves by 180

This time, the

maximum of the firstwave is at the

minimum of the

second and vice-versa, so when you

add them up, you get

zero.

When the wavesinteract so that the

sum is larger than the

original waves, wecall that constructiveinterference. When

they interact so that

the sum is smaller, wecall that destructive

interference. You

can have everything

y 1

y 2

y 1 + 2

t

t

t

y1

y2

y1+ y2t

t

t


2/8

Lecture 27 Page 2

in between partially destructive and partially constructive interference.

One of the reasons we care about how the waves interact with each other is because there are a

number of places where waves travel into an object like an organ pipe and travel out again.Use the rope as an example. If I shake the rope, a pulse traveling down the rope will reach the

fixed end and will reflect back inverted. If I keep shaking the rope, I set up a wave train such

that, when one pulse reaches the end of the line and turns around, it will interfere with one of thepulses still heading toward the wall. There will be some points along the rope where the waves

interact constructively and some points where they interact destructively. The result is that there

are some points on the rope that are always standing still. We call these nodes. There are other

points at which the wave has maximum values, which we call anti-nodes. The waves that resultfrom this are called standing waves.

If I move my hand faster up and down, you see that I can change the number of nodes and

antinodes. The length of the rope limits the configurations I can set up. The fundamental is the

configuration in which there are no nodes (except the two at the end). When you pluck a guitarstring, for example, you are exciting the fundamental. If you change the length of the string by

holding it at one of the frets, you change the wavelength and thus the frequency heard.

Some nomenclature:

Any frequency that is an integral multiple of the fundamental is called a harmonic.

The first harmonic is the frequency, which we'll denote as f1.

The second harmonic has a frequency exactly twice the fundamental, so f2 = 2f1

The first harmonic is the situation in which there is one node. Two nodes denote the secondharmonic, etc.

The other piece of nomenclature is the idea of an overtone. Overtones are the harmonics above

the fundamental frequency. The first overtone for a wave on a string is the second harmonic.

The second overtone for a wave on a string is the third harmonic

Youll notice that we dont have many options here. There are either one, two, three, etc. nodes

on our string. This limits the number of patterns we can have. Lets investigate how manypatterns are possible and the conditions under which they are produced. The chart at left shows

that there is a

pattern. The nth

harmonic is related

to the length as

Ln n

=

2

where n = 1 for the

first harmonic, 2 forthe second

harmonic, etc.

The wavelengths

for each harmonicare given by:

1 /2 = L

2 = L

3 3 /2 = L

2 4= L

F i r s t H a r m o n i c

( fu n d a m e n t a l)

S e c o n d H a r m o n i c

( f ir s t o v e r t o n e )

T h i r d H a r m o n ic

( s e co n d o v e r t o n e )

F o u r t h H a r m o n i c

( t h i r d o v e r t o n e )


3/8

Lecture 27 Page 3

Ln

L

n

n

n

=

=

2

2

The n

th

harmonic will always have n loops in the wave pattern.

Note that the frequency and the wavelength of each wave on the string is different, but that

the all the waves have the same velocity.

v = f11 = f22 = f33

and so on.

We can related the harmonics to the fundamental as follows:

fv L

n

fv

L

n

f nv

Lnf

n

n

n

n

n

= =

=

=FHG

IKJ

=

;substitute in

2

2

21

EXAMPLE 27-1: A guitar string has a fundamental frequency of 440 Hz and a length of 0.50 m.

a) Draw the picture of the first five overtones and find their frequencies.b) What are the wavelengths of the waves on the string?c) What is the velocity of waves on the string?d) What is the velocity of the sound waves produced by the string?Solution a: The first three overtones are given by the picture above. The fourth overtone

(which is the same as the fifth harmonic) will have four nodes/five loops. The fifth overtone(which is the same as the sixth harmonic) will have five nodes and six loops

A harmonic is an integral multiple of the fundamental. We will always have that fn = nf1, so

f f Hz Hz

f f Hz Hzf f Hz Hz

f f Hz Hz

f f Hz Hz

2 1

3 1

4 1

5 1

6 1

2 2 440 880

3 3 440 13204 4 440 1760

5 5 440 2200

6 6 440 2640

= = =

= = =

= = =

= = =

= = =

b g

b gb g

b g

b g

Notice that the difference between any two harmonics that differ by one will always be equal to

the fundamental frequency.


4/8

Lecture 27 Page 4

f4 f3 = 4f1 3f1 = f1

f3 f2 = 3f1 2f1 = f1

Solution b: The wavelength of the waves can be found from

n

L

nL

m m

Lm

Lm m

Lm m

Lm m

=

= = =

= =

= = =

= = =

= = =

2

2

12 050 10

2

20 50

2

3

2

3050 0 33

2

4

1

20 50 0 25

2

5

2

5 050 0 20

1

2

3

4

5

( . ) .

.

. .

. .

. .

Solution c: The velocity of waves on the string is given by

v f Hz m ms

= = =1 1 440 100 440 b gb g.

Note that you get the same thing if you multiply any fn andn!

Solution d: The velocity of the sound waves produced will be 340 m/s, which is the general

speed of sound at 15C. Don't confuse the two velocities!

EXAMPLE 27-2 A nylon string is stretched between supports 1.20 m apart.

a)

what is the wavelength of waves on this string?b) If the speed of transverse waves on a string is 850 m/s, what is the frequency of the first

harmonic and the first two overtones?

a) To determine the wavelength, draw thefundamental.

The fundamental is one half of a wavelength. The

wavelength is therefore twice the length of thestring, or 2.40 m.

b) The frequency of the first harmonic is given byf

v

fm

Hzms

1

1

1

850

2 40350

=

= =

.

The frequency of the first two overtones are given by f2 = 2f1 = 700 Hz and f3 = 3f1 = 1050 Hz

1.20 m


5/8

Lecture 27 Page 5

Waves in Tubes

String instruments produce sound by causingvibrations in the string. These vibrations excite the

air around the string, causing the air around the

strings to be alternately compressed and rarefied

creating sound waves. Note that the velocity ofwaves on a string is not the same thing as the

velocity of sound waves.

In wind instruments, the pressure variations arecontrolled by using a column. Consider a tube of

length L that is open at both ends. When you blow

into the tube, you create a longitudinal wave. Thesound wave is thus created directly. In a string

instrument, you create a transverse wave on the

string, which then

excites the airsurrounding the string

and creates the sound

wave (which islongitudinal). The

sound wave is created

directly by windinstruments.

Longitudinal waves

are variations in the

density of the air in a

given part of the tube. If you set up longitudinal standing waves, we find an analogous situationto the transverse standing waves seen on a string. If we could take a picture of the movement of

the air molecules in each part of the standing wave, we would find the following: at some pointsalong the standing wave, there is no motion of the molecules at that position. This is called

displacement node, exactly like the node along a string when the string doesnt move. Similarly,

there are points along the tube where the molecules oscillate at their maximum amplitudes.

These are displacement antinodes. We can plot the amplitude of the motion of the molecules toillustrate this. Note that the diagrams for the production of sound by wind instruments are

different, because we're plotting displacement waves and not the actual shape of the air.

Waves in a pipe open at both ends.

We're going to be working in the limit of the tube length being much greater than the diameter of

the tube. This allows us to ignore effects at the ends of the tube that would complicate ourdescription.

At the open end of a column of air, the air molecules can move freely, so there will be a

displacement antinode at the open end of a pipe. We can use the same approach to determine the

modes of a tube of length L open at both ends are as we did in finding the waves on a string -draw the possibilities.

d i s p l a ce m e n ta n t i n o d e

d i s p la c e m e n t

n o d e

Making Sound with Strings

Ear


6/8

Lecture 27 Page 6

For the first harmonic (fundamental), wehave half of a wavelength in the tube

L =

2

For the second mode, we have a full

wavelength in the tube

L = =2

2

For the third mode,

L =3

2

so in general, we can extrapolate this to:

L n

f nv

L

n

n

=

=

2

2

These formulas are good for waves in a tube open on both ends.

Examples of instruments with pipes open at both ends:

flute trumpet organ pipes

You change the length of the tube by pressing keys. In a flute, closing a key elongates the tube.

In a trumpet or French horn, pressing keys adds additional lengths of tubing to the pipe.

EXAMPLE 27-3: a) Calculate the fundamental frequency and the first three overtones of ahollow pipe open at both ends having length 30.0 cm. b) Calculate the wavelength of each wave.

We have

f nv

Ln =

2

so

fv

L

fm

f Hz

ms

1

1

1

2

340

2 0 30

570

=

=

=

.b g

1/2 = L

2 = L

3 3/2 = L

2 4 = L

F i r s t H a r m o n i c

S e c o n d H a r m o n i c

T h i r d H a r m o n i c

F o u r t h H a r m o n i c

f1 = v / 1 = v / 2 L

f2 = v / 2= v / L

= 2 v / 2 L

f3 = v / 3= 3 v / 2 L

f4 = v / 4= 2 v / L

= 4 v / 2 L


7/8

Lecture 27 Page 7

We have the same relationship between the frequencies: fn = nf1. So f2 = 2f1 = 1140 Hz and f3 =

3f1 = 1710 Hz.

b) The wavelength is given byv f

vf

Hzm

Hzm

Hzm

n n

n

n

ms

ms

ms

=

=

= =

= =

= =

1

2

3

340

5700 60

340

11400 30

340

15700 22

.

.

.

Standing Waves in a Pipe Open on One End

We can also have pipes that are closed on one end and open on the other. (Closed on two ends

wouldnt make any sense.) This is a slightly different case, because at the closed end, were

required to have a displacement node, which will change the wave patterns allowed. Althoughthe frequencies of waves in a pipe open at two ends are the same as those of a string with the

same length, the case of waves in a pipe open at only one end will be quite different. We candraw the allowed patterns as shown.

In general,

L n n=

4

f nv

Ln =

4

but n can only be odd! Therefore, we

talk about this case having only odd

harmonics. There are only 1, 3, 5We

call 3 the first overtone, 5 the second

overtone, etc.

Examples of instruments with pipes closed

at one end include organ pipes

EXAMPLE 36-4: a) Calculate the

fundamental frequency and the first three

overtones of a hollow pipe open at oneend having length 30.0 cm. b) Calculate the wavelength of each wave.

1 /4 = L

L = 3 3/4

f1 = v / 1 = v / 4 L

f 3 = v / 3 = 3 v / 4 L

f5 = v / 5 = 5 v / 4 LL = 5 5/4

f7 = v / 7 = 7 v / 4 LL = 7 7/4

W a v e s in P i p e s O p e n a t O n e E n d


8/8

Lecture 27 Page 8

We have

f nv

Ln =

4, but we are restricted to n odd

so

( )

( )

( )

( )

1

3

5

7

340283

4 4 0.30

3403 3 850

4 4 0.30

3405 5 1420

4 4 0.30

3407 7 1980

4 4 0.30

ms

ms

ms

ms

vf Hz

L m

vf Hz

L m

vf Hz

L m

vf Hz

L m

= = =

= = =

= = =

= = =

c) The wavelength is given byv f

v

f

Hzm

Hzm

Hzm

Hzm

n n

n

n

ms

ms

ms

ms

=

=

= =

= =

= =

= =

1

3

5

5

340

283120

340850

0 40

340

14200 24

340

1980018

.

.

.

.

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